My problem originates from the following expression (and a slight lack of deeper knowledge in probability theory):
$\mathbb{E}[\exp{(8\theta Y_{t})}]$, where
$$Y_{t}:=\int_{0}^{t}\int_{0}^{s}\frac{W_{s}-W_{r}-y}{\vert W_{s}-W_{r}-y\vert^{2}}\:dr\:dW_{s}$$
and $W_{s}$ is a 2-dimensional Brownian motion.
My goal is to show that (for $ \theta < \theta_{0}$) it is finite. I already managed to show that if one is able to proof this for the quadratic variation of the term in the exponent, one obtains the existence of such a $\theta_{0}$. So I need to show the following:$\mathbb{E}[\exp{(\gamma C\langle Y_{t} \rangle)}]<\infty$ for some $\gamma$. If my estimates are correct, $\mathbb{E}\left[\exp{\left(\gamma C \sum_{i=1}^{2}\int_{0}^{t}\left(\int_{0}^{s}\frac{(W_{s}-W_{r}-y)_{i}}{\vert W_{s}-W_{r}-y\vert^{2}}\:dr\right)^{2}\:ds_{i} \right)}\right]\leq \mathbb{E}\left[\exp{\left( C\gamma\int_{0}^{t}\left(\int_{0}^{t}\frac{1}{\vert W_{s}-W_{s-r}-y\vert}\:dr\right)^{2}\:ds\right)}\right]$
$\leq \mathbb{E}\left[\exp\left(\gamma Ct\left(\int_{0}^{t}\frac{1}{\vert W_{r}-y\vert}\:dr\right)^{2}\right)\right]$.
I already found out, that I could use Ito's formula, applied to a 2-dimensional Bessel process($X_{t}$), to reformulate the term in the integral :$\frac{1}{2}\left(\int_{0}^{t}\frac{1}{\vert W_{r}-y\vert}\:dr\right)=X_{t}-\vert y\vert-B_{t}$, where $B_{t}:=\frac{\sum_{i=1}^{2}W^{i}_{t}\:dW^{i}_{t}}{\vert W_{t}\vert}$ is a 1-dimensional Brownian motion.
This is the point where I am stuck, as I don't see an easy way to evaluate the square of this expression.